You are not logged in.

- Topics: Active | Unanswered

**razor****Member**- Registered: 2006-02-24
- Posts: 6

We have the real positive(only) numbers space of dimension n.We have also a relation of two members(sorry),continuous,absoletuly monotonic(only >),cursive,in this space.

Also x,y are arrays and x>y in this space and 0<a<1.

How can we proove that a*x+(1-a)*y>y ?. (*: multiplication)

(G. Aliprantis, D. Brown and O. Burhinshaw, Existence and Optimality of Competitive Equilibrium, Springer-Verlag, 1990.)

sorry for my english.

*Last edited by razor (2006-03-05 10:59:29)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

ax + (1-a)y > ay + (1-a)y = y(a + 1 - a) = y

ax + (1-a)y > y

All that needed was x > y.

*Last edited by Ricky (2006-03-05 12:56:13)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline